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Semenov [28]
3 years ago
12

A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj

ectory, and then pushed forward with an initial speed of 2.93 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?
96 J

136 J

265 J

180 J
Physics
1 answer:
Alika [10]3 years ago
4 0

Answer:

265 J

Explanation:

Energy=PE+KE=mgh+ 0.5mv^{2} where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J

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A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

  • m = 10 kg
  • g = 9.8 m/s²
  • T = ?

Use formula:

  • \boxed{\bold{T=m*g}}

Replace and solve:

  • \boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}
  • \boxed{\boxed{\bold{T=98\ N}}}

The tension in the rope is <u>98 Newtons.</u>

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3 years ago
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RUDIKE [14]

Answer:

I think deposition of C is the oldest deposit

8 0
3 years ago
A satellite is in circular orbit around the earth. The orbit the satellite is at a height of 420 km above the earth's surface. F
astraxan [27]

Answer:

7661.06 m/s

Explanation:

R = Radius of Earth = 6.38\times 10^6\ m

h = Distance from the Earth = 420000 m

G = Gravitational constant = 6.674\times 10^{-11} N m^2/kg^2

M = Mass of Earth = 5.98\times 10^{24}\ kg

V=\sqrt{g{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{GM}{R^2}{\frac{R^2}{R + h}}}\\\Rightarrow V=\sqrt{\frac{6.674\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^6)^2}{\frac{(6.38\times 10^6)^2}{6.38\times 10^6 + 420000}}}\\\Rightarrow V=7661.06\ m/s

The orbital speed of the satellite is 7661.06 m/s

6 0
4 years ago
Describe an experiment you could perform to determine the average speed of a toy car rolling down an incline
Anna35 [415]
Imagine an incline of length "x". You let the toy go at the top of the incline and measure the time "t" it takes for it to get to the bottom.
The average speed would be "v=x/t".
8 0
3 years ago
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa
OLga [1]

Answer:

<h2>15m/s</h2>

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − \omegat) where An is the amplitude f oscillation, \omega is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f where;

\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = \frac{1}{(2/15)}

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength \lambda = 2m

Transverse speed v = f \lambda

v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s

Hence, the transverse speed at that point is  15m/s

8 0
3 years ago
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